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3 years ago

What is the perimeter of this triangle

Mathematics

1 answer:

DENIUS [597]3 years ago

3 0

Answer:

they are all equal side so add them up

Step-by-step explanation:

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What is the slope of (-5,-1) (5,5)? Explain.

ArbitrLikvidat [17]

Step-by-step explanation:

$slope = \frac{ - 1 - 5}{ - 5 - 5} = \frac{ - 6}{ - 10} = \frac{ 3}{ 5} \\$

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3 years ago

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If a kilowatt-hour of energy costs 75 cents, how much does it cost to leave a 40-watt light bulb on for the entire month of May

spin [16.1K]

May has 31 days
31 x 24 = 744 hours in may
744 x 40 = 29760 watt hours
29760 watt hours = 29.76 kwh
29.76 x 75 = 2232 cents
2232 cents = $22.32

8 0

3 years ago

Bridgette bought 1 pack of white T-shirts and 5 packs of blue T-shirts for her basketball team. The white T-shirts come in packs

irakobra [83]

<h3>Answer: 32</h3>

Work Shown:

1 pack of white = 2 shirts = A

1 pack of blue = 6 shirts

5 packs of blue = 5*6 = 30 shirts = B

A+B = 2 white shirts + 30 blue shirts = 32 shirts total

5 0

3 years ago

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What has a foot on each end and in the middle

cricket20 [7]

Is it a yardstick? It has 2 ft of the edges and one ft in the middle.

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3 years ago

I need to know the improper fractions answers for:

zmey [24]

Answer:

Part 1) $x=\frac{15}{2}\ units$

Part 2) $z=\frac{15\sqrt{3}}{2}\ units$

Part 3) $y= \frac{15\sqrt{3}}{4}\ units$

Part 4) $b= \frac{45}{4}\ units$

Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

$cos(60^o)=\frac{x}{15}$ ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

$cos(60^o)=\frac{1}{2}$

substitute

$\frac{1}{2}=\frac{x}{15}$

solve for x

$x=\frac{15}{2}\ units$ ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

$15^2=x^2+z^2$

substitute the value of x

$15^2=(\frac{15}{2})^2+z^2$

solve for z

$z^2=15^2-(\frac{15}{2})^2$

$z^2=225-\frac{225}{4}$

$z^2=\frac{675}{4}$

$z=\frac{\sqrt{675}}{2}\ units$

simplify

$z=\frac{15\sqrt{3}}{2}\ units$

step 3

Find the value of y

In the right triangle of the right

$sin(30^o)=\frac{y}{z}$ ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$sin(30^o)=\frac{1}{2}$

$\frac{1}{2}=y:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{1}{2}= \frac{2y}{15\sqrt{3}}$

$y= \frac{15\sqrt{3}}{4}\ units$

step 4

Find the value of b

In the right triangle of the right

$cos(30^o)=\frac{b}{z}$ ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}$

solve for y

$\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}$

$b= \frac{45}{4}\ units$

7 0

3 years ago